#coulombs calculation

Variation of Conductivity with Concentration and Measurement

How many coulombs are required to deposit 0.27 g of aluminum from Al(NO₃)₃ solution? (Atomic mass of Al = 27 g/mol, F =

Al³⁺ + 3e⁻ → Al . 1 mol Al (27 g) requires 3F. Moles = (0.27/27) = 0.01 mol , Charge = 0.01 × 3 × 96500 = 2895 C .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement

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Electrolytic Cells and Electrolysis and Faraday's Laws

During electrolysis of aqueous NaI with inert electrodes, 0.224 L of H₂ (STP) is produced at the cathode. How many coulo

Cathode: 2H₂O + 2e⁻ → H₂ + 2OH⁻ . Moles = (0.224/22.4) = 0.01 mol , Charge = 0.01 × 2 × 96500 = 1930 C .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrolytic Cells and Electrolysis and Faraday's Laws

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Electrochemical Cells - Galvanic Cells and Electrode Potential

How many coulombs are required to reduce 1 mole of Br₂ to Br⁻ ? (F = 96500 C/mol)

Br₂ + 2e⁻ → 2Br⁻ . 1 mol Br₂ requires 2F. Charge = 2 × 96500 = 193000 C .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

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Electrochemical Cells - Galvanic Cells and Electrode Potential

How many coulombs are required to reduce 1 mole of Br₂ to Br⁻ ? (F = 96500 C/mol)

Br₂ + 2e⁻ → 2Br⁻ . 1 mol Br₂ requires 2F. Charge = 2 × 96500 = 193000 C .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

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Batteries - Primary Secondary and Fuel Cells

In a dry cell, 0.435 g of MnO₂ (molar mass 87 g/mol) is reduced to MnO(OH). How many coulombs are consumed? (F = 96500 C

MnO₂ + H⁺ + e⁻ → MnO(OH) . Moles = (0.435/87) = 0.005 mol , Charge = 0.005 × 96500 = 482.5 C .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells

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Batteries - Primary Secondary and Fuel Cells

How many coulombs are required to deposit 5.4 g of aluminum from an AlCl₃ solution? (Atomic mass of Al = 27 g/mol, F = 9

Al³⁺ + 3e⁻ → Al . 1 mol Al (27 g) requires 3F. Moles = (5.4/27) = 0.2 mol , Charge = 0.2 × 3 × 96500 = 57900 C .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells

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Batteries - Primary Secondary and Fuel Cells

How many coulombs are required to deposit 5.4 g of aluminum from an AlCl₃ solution? (Atomic mass of Al = 27 g/mol, F = 9

Al³⁺ + 3e⁻ → Al . 1 mol Al (27 g) requires 3F. Moles = (5.4/27) = 0.2 mol , Charge = 0.2 × 3 × 96500 = 57900 C .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells

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Batteries - Primary Secondary and Fuel Cells

How many coulombs are required to deposit 108 g of silver from an AgNO₃ solution? (Atomic mass of Ag = 108 g/mol, F = 96

Ag⁺ + e⁻ → Ag . 1 mol Ag (108 g) requires 1F. Moles = (108/108) = 1 mol , Charge = 1 × 96500 = 96500 C .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells

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Batteries - Primary Secondary and Fuel Cells

How many coulombs are required to deposit 108 g of silver from an AgNO₃ solution? (Atomic mass of Ag = 108 g/mol, F = 96

Ag⁺ + e⁻ → Ag . 1 mol Ag (108 g) requires 1F. Moles = (108/108) = 1 mol , Charge = 1 × 96500 = 96500 C .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells

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